Step-by-step solved example
Cement 564 lb (SG=3.15), water 282 lb (SG=1), no air. Volume of these two?
1. Cement
V_c = 564/(62.4×3.15) = 2.87 ft³.
2. Water
V_w = 282/62.4 = 4.52 ft³. Sum = 7.39 ft³.
Answer: 7.39 ft³ (cement+water)
Cement 564 lb (SG=3.15), water 282 lb (SG=1), no air. Volume of these two?
1. Cement
V_c = 564/(62.4×3.15) = 2.87 ft³.
2. Water
V_w = 282/62.4 = 4.52 ft³. Sum = 7.39 ft³.
Answer: 7.39 ft³ (cement+water)
1.γw in US customary is about
2.1 yd³ =
3.SG of portland cement is about
4.5% air in a 27 ft³ batch is
5.Over-yield means actual volume is
6.SSD aggregate weights already include
7.W = 312 lb water, γw = 62.4. V_w =
8.If yield is 26 ft³ instead of 27, quantities per yd³ will be
9.Absolute volume method is also used in
10.Mass / SG / γw must use